
#1
Dec1003, 09:34 AM

P: 661

Here is the Problem:
[tex] \int_0^\pi\theta^2cosec^2\theta d\theta[/tex] I want to do this problem by parts and finding hard to find shortest way to do the problem any shortest way pls 



#2
Dec1003, 04:45 PM

Sci Advisor
P: 5,935

The straightforward way is probably the easiest. integral of csc^{2} is ctn and integral of ctn is log(sin). Integrating log(sin) may take a little work.




#3
Dec1003, 05:05 PM

P: 640





#4
Dec1003, 05:24 PM

P: 678

integral of csc2 is ctn 



#5
Dec1003, 09:49 PM

P: 661

It is cosecant(x) i.e csc(x)
There are many ways of doing a problem i am looking for shortest way Thnxs 



#6
Dec1003, 10:39 PM

P: 678

How about:
[tex] \begin{align*} \int_0^\pi\theta^2\csc^2\theta\;d\theta &=\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\theta^2\csc^2\theta\ ;d\theta \\ &>\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\csc^2\theta\;d\th eta \\ &=\int_0^1\theta^2\csc^2\theta\;d\theta+\left[\cot\theta\right]_1^\pi \end{align*} [/tex] Now since [itex]\lim\limits_{\theta\rightarrow\pi}(\cot\theta)=+\infty[/itex] we know that the rightmost term diverges. Moreover, the left integral (the one from 0 to 1) is clearly positive. Thus the original integral clearly diverges. 



#7
Dec1003, 11:31 PM

P: 3

master_coda:
the integral of x^2*(cscx)^2 from 1 to pi comes out to be about 4.2 the integral of (cscx)^2 from 1 to pi goes to inifinite. I believe to say the integral of x^2*(cscx)^2 from 1 to pi is greater than the integral of (cscx)^2 from 1 to pi since according to the values it came out as then the formula should be switched around and divergence would not be proved. himanshu121: Best way to do it would be to plug it into a calculator or look it up in a table. Otherwise you're probably stuck with integrating log(sin(x)). jk 



#8
Dec1103, 12:35 AM

P: 661

I found the way but dont know whether it is shortest one or not but definetly i wont stuck at log(sinx) [tex] I = \int_0^\pi\theta^2\csc^2\theta d\theta [/tex] [tex]I= \int_0^\pi\ (\pi\theta)^2\csc^2\theta d\theta [/tex] this gives [tex] \pi\int_0^\pi\csc^2 \theta d\theta = 2\int_0^\pi\theta\csc^2\theta d\theta [/tex] integrating by parts with one part as [tex]\theta[/tex] and other as [tex]\theta\csc^2\theta d\theta [/tex] i will get[tex] \int_0^\pi \cot\theta d\theta = \log(csc\thetacot\theta) [/tex] much easier than integrating log(sinx) But another problem is how i will put the limits in cotx from 0 to pi in both cases it is infinity and i know there is no break in the function cotx b/w these points 



#9
Dec1103, 07:09 AM

Emeritus
Sci Advisor
PF Gold
P: 16,101

I agree with master_coda...




#10
Dec1103, 09:44 AM

Emeritus
Sci Advisor
PF Gold
P: 10,424

 Warren 



#11
Dec1103, 09:45 AM

P: 678

I'm pretty sure that the integral diverges. My algebra, my calculator and my computer all agree.
Also, if it's any help, [itex]\csc\theta\cot\theta=\frac{1\cos\theta}{\sin\theta}[/itex]. 



#12
Dec1303, 09:02 PM

P: 3

hurkyl. yeah that 4.2 was kinda off huh? just keeping you on your toes i guess:P i entered it in wrong but i redid it and came up with what you got.
i used u=(x^2)csc^2(x) and dv=dx and came up with the first term (x^3*csc^2(x)) going to infinite so i think you might be right coda. jk 



#13
Dec1703, 03:54 PM

P: 76

Why dont you try using the tabular method to doing this problem.
Since theta^2 will eventually go to zero if you keep on taking the derivatives, you should do it by tabular method 


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